1.直接计算2x2占用的屏幕数 2.计算已占用屏幕的情况下的剩余空间 3.如果剩余空间足以放下1x1，屏幕数为1中屏幕数 4.如果不足，则1x1数减去2中剩余空间，再求剩下的1x1占格子数

#include <iostream>
using namespace std;

int main() {
    int n;
    int a,b;
    cin>>n;
    while (~scanf("%d %d",&a,&b)){
        int screens_for_b = (b + 1) / 2;
        int remain = screens_for_b * 15 - b * 4;
        if (remain>a){
            cout<<screens_for_b<<endl;
        }else{
            cout<<screens_for_b+(a-remain+14)/15<<endl;
        }
    }
    return 0;
}